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Q 12.7 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels

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As per Bohr's model the angular momentum of electrons in each orbit is constant and a multiple of \frac{nh}{2\pi }

m_{e}v_{n}r_{n}=\frac{nh}{2\pi }           (i)

The electrostatic force of attraction between the electron and the nucleus provides the required centripetal force for the circular motion of the electron.

\frac{mv_{n}^{2}}{r_{n}}=\frac{e^{2}}{4\pi \varepsilon _{0}r_{n}^{2} }       (ii)

Using equation (i) and (ii) we get

v_{n}=\frac{e^{2}}{2nh\varepsilon _{_{0}}}

r_{n}=\frac{n^{2}h^{2}\varepsilon _{_{0}}}{m_{e}\pi e^{2}}

\\v_{1}=\frac{e^{2}}{2h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{2\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}

v1=2.18\times106 ms-1

\\v_{2}=\frac{e^{2}}{4h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{4\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}

v2=1.09\times106 ms-1

\\v_{3}=\frac{e^{2}}{6h\varepsilon _{_{0}}}\\ \\v_{1}=\frac{(1.6\times 10^{-19})^{2}}{6\times 6.62\times 10^{-34}\times 8.85\times 10^{-12}}

v3=7.28\times105 ms-1

 

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